
\prob{0097}{三元二次方程组II}

求方程组
\[ \left\{\begin{aligned}
  6xy &= 2x + 5y \\
  -2yz &= 3y + 2z \\
  -zx &= 5z + 6x
\end{aligned}\right. \]
的所有解。
\problabels{yellow/代数, green/方程相关问题}

\ans{$x = y = z = 0$或$x = 1, y = 2, z = -1$}

\subsection{升次消项}

由原方程组知
\begin{align}
  6xyz &= 5yz + 2zx \label{eq:0097-52} \\
  -6xyz &= 9xy + 6zx \label{eq:0097-96} \\
  -6xyz &= 30yz + 36zx \label{eq:0097-3036}
\end{align}
式~\ref{eq:0097-52} 分别与式~\ref{eq:0097-96}、式~\ref{eq:0097-3036} 相加，消去三次项，得
\begin{align*}
  9xy + 5yz + 8zx &= 0 \\
  36xy + 35yz + 2zx &= 0
\end{align*}
消去$xy, yz$项，得
\[ yz = 2zx, xy = -2xz \]
若$z = 0$或$x = 0$，则$x = y = z = 0$，是方程组的一个解；若$z \ne 0$且$x \ne 0$，则
\[ y = 2x, y = -2z, z = -x \]
代入$-zx = 5z + 6x$知
\[ x^2 = x, x \ne 0 \Rightarrow x = 1 \]
于是有$x = 1, y = 2, z = -1$。故方程所有解为
\[ \left\{ \begin{aligned}
  x &= 1 \\ y &= 2 \\ z &= -1
\end{aligned} \right. \text{或} \left\{ \begin{aligned}
  x &= 0 \\ y &= 0 \\ z &= 0
\end{aligned} \right. \]
